比较 R 中回归的两个斜率

Question

I used this code to get a plot with log formula

df %>%
  mutate(absVO2 = absVO2*1000) %>%
  ggplot(aes(x = VE, y = absVO2, color = group)) +
  geom_point(size = 0.5) +
  geom_smooth(aes(colour=group), method = "lm", formula = y ~ log(x)) +
  stat_regline_equation(label.x = 25, label.y = c(2400,2800), 
                        formula = y ~ log(x),
                        aes(color = group, label =  paste(..eq.label.., ..adj.rr.label.., sep = "~~~~")),) +
  xlab("VE (L/min)") + 
  ylab(expression(paste("V", O[2]," (mL/min)"))) +
  scale_x_continuous(limits = c(0, 100), breaks = seq(0, 100, by = 25)) +
  theme_classic() +
  theme(
    panel.background = element_rect(fill ="white", colour = "white"),
    axis.line = element_line(colour = "black"),
    legend.title = element_text(face = "bold"),
    legend.position = "none",
    strip.text = element_blank()
  )

That gives me this output:

I would like to statistically compare the two slopes, ie 910 and 970 which are the a in the VO2 = a * log(VE) + b formula.

Thank you!

Here are the data:

structure(list(VE = c(22.75, 23.75, 26.75, 32.5, 39.5, 48.25, 
55.75, 24, 26.5, 29, 36.5, 44.75, 55.25, 63.75, 78, 19, 20.75, 
25, 29, 39.25, 51, 25, 25.5, 28.5, 34.25, 35.5, 40.25, 47, 52.5, 
59.5, 64.5, 83, 92.25, 18.25, 19.75, 20.75, 23.5, 26.25, 28, 
32.75, 37.25, 24.25, 26.75, 29.5, 31, 32, 39, 43.25, 46.25, 51.75, 
52.75, 64.5, 72.75, 15.75, 16.25, 20.25, 28, 34.75, 45.75, 64.75, 
18.25, 21.25, 22.25, 24.75, 29.5, 21.25, 20.75, 26, 30.75, 34, 
38.5, 46.75, 53.75, 70.5, 82.75, 98.25, 17.25, 22, 25.75, 29.75, 
33.75, 38.75, 44.5, 50.75, 56.25, 65, 71.75, 81, 99, 30, 29, 
29.75, 38.25, 40.75, 47.25, 52.75, 65.5, 22.75, 24.75, 28.75, 
34, 48, 53.75, 61.5, 73.75, 22, 21.75, 24.25, 28, 34.5, 40, 48.5
), absVO2 = c(0.739, 0.81975, 0.95125, 1.07525, 1.199, 1.34575, 
1.49775, 0.66125, 0.7485, 0.855, 0.9845, 1.1495, 1.3465, 1.451, 
1.5985, 0.61675, 0.717, 0.86275, 0.98575, 1.13, 1.262, 0.8835, 
0.94575, 1.08125, 1.244, 1.30475, 1.4735, 1.679, 1.79075, 1.96, 
2.0405, 2.34425, 2.4435, 0.5925, 0.661, 0.7435, 0.87875, 0.9435, 
0.99675, 1.11425, 1.20275, 0.9255, 1.02925, 1.11675, 1.19725, 
1.24775, 1.42625, 1.54225, 1.59425, 1.69675, 1.776, 1.94525, 
2.0395, 0.535, 0.5845, 0.76875, 0.982, 1.09975, 1.259, 1.3265, 
0.7205, 0.86825, 0.9235, 1.01075, 1.17275, 0.7435, 0.76575, 0.96075, 
1.0975, 1.21975, 1.34525, 1.50625, 1.6755, 1.86325, 2.0465, 2.1395, 
0.613, 0.85525, 0.98, 1.076, 1.287, 1.4615, 1.59325, 1.6965, 
1.884, 1.998, 2.1425, 2.31275, 2.474, 0.81075, 0.8035, 0.8645, 
1.076, 1.119, 1.255, 1.33825, 1.499, 0.8105, 0.8865, 1.03725, 
1.134, 1.26675, 1.36275, 1.4455, 1.52875, 0.64675, 0.6645, 0.742, 
0.81675, 0.983, 1.06875, 1.155), group = c("CAD", "CAD", "CAD", 
"CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", 
"CAD", "CAD", "CAD", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "Healthy", "CAD", "CAD", "CAD", 
"CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", 
"Healthy", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", 
"CAD", "CAD", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "CAD", "CAD", "CAD", "CAD", "CAD", "CAD", 
"CAD", "CAD", "Healthy", "Healthy", "Healthy", "Healthy", "Healthy", 
"Healthy", "Healthy", "Healthy", "CAD", "CAD", "CAD", "CAD", 
"CAD", "CAD", "CAD")), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -112L))

Answer 1

Your hypothesis is that the slope of two groups are different. To answer this you build a linear model with added group variable and an interaction between group and VE (without an interaction you are assuming the slope is the same between the groups, so you already have your answer in this case):

> summary(lm(absVO2*1000~group*log(VE),data=df))

Coefficients:
                     Estimate Std. Error t value Pr(>|t|)    
(Intercept)          -2120.98     210.38 -10.081   <2e-16 ***
groupHealthy           -96.46     263.82  -0.366    0.715    
log(VE)                914.32      57.40  15.928   <2e-16 ***
groupHealthy:log(VE)    60.52      72.64   0.833    0.407    

Intercept is the value of absVO2 when log(VE) is zero (not really meaningfull for your data) and group = CAD (which is the reference group, you can change this). log(VE) is the term you are interested in (the slope), which is approx. 914 for group = CAD , for group = healthy you have to add approx. 60 to it (which would result in a slope of approx. 974), but this coefficient is not significant (based on the PR(t) column - the dreaded p-value).